Second Law violations in the wake of the Electrocaloric Effect in liquid dielectrics Andreas Trupp, Fachhochschule der Polizei des Landes Brandenburg -University of Applied Science- Private e-mail: [email protected] A short version of this article was published in: AIP Conference Proceedings Vol. 643: Quantum Limits to the Second Law, First International Conference on Quantum Limits to the Second Law, University of San Diego, California, 28-31 July 2002 Abstract: In any textbook on physics, Coulomb’s law of the mutual force between two point charges q0 at a distance r is modified by the appearance of the term K if the point charges are embedded in a dielectric: The dimensionless constant K ( 1) denotes the permittivity of the dielectric. According to this formula, the force is either reduced by the factor 1/K -if the charges q0 are kept invariant in amount-, or is increased by the factor K -if the potential V is kept invariant- as a result of the introduction of the dielectric. V is the potential of the location of one point charge as a result of the field generated by the other point charge. Feynman argues that the formula is correct only if the dielectric is a liquid, and that it does not work properly with solids. His criticism does not go far enough. Two simple experiments with a liquid dielectric (backed by theoretical reflections) reveal that the formula is correct only if the two point charges have opposite signs (negative and positive). If the signs are equal, the formula reads (when applied to point charges in liquid dielectrics): Hence the force is reduced by the factor 1/K2 if the charges (of equal sign) are kept invariant, and is left unaffected by the introduction of the dielectric if the potential V is kept invariant. With a so revised formula, cyclic processes can be performed in which the electrocaloric effect (that heats up the dielectric when the electric field is being built, and cools down the dielectric when the field is disappearing) is no longer symmetrical, leading to the conversion of ambient heat to electric work as a net result of the work cycle. 1. Qualitative description of the Second Law violation It is a well established fact that internal energy in the form of heat, taken from a single reservoir (the ambient) only, can be completely converted to mechanical work without any refuse heat (that would be an unwanted side-product). Such a conversion is, for instance, taking place when an ideal gas, concealed in a cylinder, is expanding isothermally at ambient temperature and is thereby moving a piston. The only heat flow involved is that from the ambient into the gas (to prevent its cooling off), with that heat being totally turned into mechanical work performed by the piston. Unfortunately, a perpetual motion machine of the second kind (that would convert ambient heat to work without requiring a second heat reservoir of a lower temperture for the reception of refuse heat) can nonetheless not be created thereby. For such a machine to operate, a cyclic process would 2 have to be performed. Any means of getting back to the starting point (that of the compressed gas in the cylinder at ambient temperature) would, however, be accompanied by a creation of heat received by the ambient. Moreover, the quantity of heat thus delivered to the ambient would -at least- be as large as the amount of heat that was previously taken from the ambient. With no heat taken from the ambient as a net result of the cycle, no net mechanical work is yielded as a result of the cycle (otherwise the law of conservation of energy would be violated). To put it the other way round: A perpetual motion machine of the second kind would be technically feasible if one could get back to the starting point without delivering heat to the ambient. The impossibility of a perpetual motion machine of the second kind operating with air as a work medium is hence not rooted in the impossibility of converting ambient heat to mechanical work at a ratio of 1:1, but in the incapability of compressing the gas isothermally without creating heat.1 A similar proposition holds true for electrostatics. Ambient heat can be converted to electric work when the electric field between the plates of a capacitor is disappearing, resulting in an adiabatic cooling of the dielectric material between the plates (electrocaloric effect). After the dielectric has regained its initial temperature (that of the ambient) by means of a heat flow from the ambient, ambient heat has been completely converted to electric work (performed -during the discharge- in the electric circuit the capacitor is part of) as a net result. 2 As with the gas in the cylinder, a perpetual motion machine of the second could be feasible if one managed to return to the starting-point without creating heat. Simply re-charging the capacitor would, however, not be an appropriate method, as the dielectric is warmed up by the same (now reversed) electrocaloric effect. In other words: The task to be solved consists in increasing the potential of a capacitor (that comprises a dielectric) without generating heat in the dielectric as an unwelcome side-effect. An attempt worth looking at might be the following: Consider a parallel-plate capacitor filled with vacuum (in order to avoid edge effects, that parallel-plate-capacitor shall be composed of two concentric spheres the diameters of which differ only slightly from each other). The voltage across shall be V0. The attractive force experienced by each plate shall be called F0. Let us now replace the vacuum by a dielectric with a relative permittivity K of well above unity. The voltage across the capacitor shall be kept invariant (at V0). Will there be a change in the attractive force (now called F1)? According to any textbook, the force will be increased by the factor K (permittivity of the 1 See a standard textbook like R.W. Pohl: Mechanik, Akustik und Wärmelehre, 18th edition, Berlin 1983, p. 336 (my own translation):“The isothermally working air compressor engine is a machine that converts the thermal energy received from the ambient; in an ideal situation the efficiency is 100%.. The conversion occurs while the compressed air is relieved from its pressure. Now we add as a new remark: The relief from pressure increases the entropy of the air. ... This increase in entropy is a permanent and relevant change that the work medium is subject to when isothermally yielding work.“ 2 One might tend to raise an objection to that description by arguing that all the electric work is the result of the conversion of energy stored in the electric field (between the two plates) rather than the result of the conversion of heat. However, at least in case we are using a common model dielectric in which the polarization charges are are substituted by regular charges on the surface (with no charges in the interior), that criticism cannot be convincing: With a given voltage across the capacitor, the field between the plates has the same structure and strength as if the dielectric were vacuum, and does hence contain the same amount of energy. Yet the electric work performed by the discharging capacitor can be many times greater than that of the discharging capacitor filled with vacuum. To account for this excess in electric work, the role of ambient heat is indispensable. 3 dielectric). As Feynman 3 has pointed out, this holds true only in case the dielectric is a liquid. If it is a solid material, the attractive force is counteracted by a virtual pressure exerted on the plate by the solid dielectric in very much the same way as the weight of our bodies is counteracted by the virtual pressure of the ground against our feet. But even with a liquid dielectric (in which the capacitor is immersed), the result is somewhat puzzling: Though a K times greater charge sits on each plate, that (free) charge is partly neutralized by polarization charges appearing on the surface of the liquid (in a common model dielectric there are no other charges than these in the whole dielectric). So, when increasing the mutual distance between the two plates, the same effective charge is moved as would be moved in case we had not replaced the vacuum by the liquid dielectric. With the effective charge being equal in amount to that in the vacuum-filled capacitor, one would (intuitively) expect F1 to be equal to F0. So, how is the increase in force accounted for? An explanation is revealed in fig. 1. In order to avoid errors by not giving consideration to the effect of the dielectric on itself, a tiny section of both the lower plate of the capacitor and the liquid is cut out and sealed off by vertical walls. The electrostatic force exerted on the free charge that sits on that tiny section of the lower plate is K times greater than it would be in case the dielectric were vacuum (at a given voltage V0). This is due to the fact that the external field (that is the field generated by all charges outside of that section) acting on the depicted free charge is the same as it would be in case the dielectric were vacuum, with the depicted free charge itself, however, being K times larger in amount. This result is not modified by the presence of the dielectric inside the section. The net force exerted on the column of dielectric (enclosed in the cut-out volume) by the external field is zero: As the external field inside that volume is homogeneous, all the dipoles (the dielectric is made up of) experience a torque only, and no translational force.